delta T = ikfm
delta T is the amount f.p. is lowered.
i is the number of particles; i.e., Na3PO4 will have i = 4 (3 for Na and 1 for PO4).
m is molality.
k is a constant and need not enter into the calculations.
Arrange the following solutions in order of decreasing freezing point.
0.20 m Na3PO4
0.35 m NaCl
0.30 m MgCl2
0.15 m C6H12O6
0.05 m CH3COOH
So, start with the first part. If i=4 and m=0.2, then delta T = 0.8. Do that for all of them and write from biggest to smallest.
actually,
delta T = iKm = 4*K*0.2 = 0.8K (you omitted the K. Although you will not need to use K, delta T is 0.8*K and not 0.8 degrees.)
You want the items to be arranged in DECREASING f.p. and since delta T is original f.p. - new f.p., then delta T will the lowest f.p. when it is the largest. So it seems to me you want to go from the smallest delta T to the largest delta T. Check my thinking.
ur stupid
Oh, I see. So we want to arrange the solutions from the smallest change in freezing point (delta T) to the largest change. Let's calculate the delta T for each solution:
0.20 m Na3PO4: delta T = 4 * 0.20 = 0.80K
0.35 m NaCl: delta T = 2 * 0.35 = 0.70K
0.30 m MgCl2: delta T = 3 * 0.30 = 0.90K
0.15 m C6H12O6: delta T = 1 * 0.15 = 0.15K
0.05 m CH3COOH: delta T = 1 * 0.05 = 0.05K
So, arranging them from smallest to largest change in freezing point, we have:
0.05 m CH3COOH
0.15 m C6H12O6
0.35 m NaCl
0.20 m Na3PO4
0.30 m MgCl2
Hope that helps!
Yes, you are correct. We want to arrange the solutions in order of decreasing freezing point, which means we need to arrange them in increasing values of delta T. Therefore, we should start with the smallest delta T and move to the largest delta T.
Let's calculate delta T for each solution:
1. For 0.20 m Na3PO4, delta T = 4 * K * 0.20 = 0.8K
2. For 0.35 m NaCl, delta T = 2 * K * 0.35 = 0.7K
3. For 0.30 m MgCl2, delta T = 3 * K * 0.30 = 0.9K
4. For 0.15 m C6H12O6, delta T = 1 * K * 0.15 = 0.15K
5. For 0.05 m CH3COOH, delta T = 1 * K * 0.05 = 0.05K
So, arranging them in increasing order of delta T, we get:
0.05 m CH3COOH
0.15 m C6H12O6
0.35 m NaCl
0.20 m Na3PO4
0.30 m MgCl2
Yes, you're correct. We want to arrange the solutions in decreasing order of freezing point, which means arranging them in increasing order of delta T values. Since delta T is given by delta T = iKm, where i is the number of particles, K is a constant, and m is the molality, we can calculate the delta T for each solution and arrange them accordingly.
Let's calculate the delta T for each solution:
- For 0.20 m Na3PO4, i = 4 and m = 0.20. So, delta T = 4 * K * 0.20 = 0.8K.
- For 0.35 m NaCl, i = 2 (since it dissociates into Na+ and Cl- ions) and m = 0.35. So, delta T = 2 * K * 0.35 = 0.7K.
- For 0.30 m MgCl2, i = 3 (since it dissociates into Mg2+ and 2 Cl- ions) and m = 0.30. So, delta T = 3 * K * 0.30 = 0.9K.
- For 0.15 m C6H12O6 (glucose), i = 1 (since it does not dissociate) and m = 0.15. So, delta T = 1 * K * 0.15 = 0.15K.
- For 0.05 m CH3COOH (acetic acid), i = 1 (since it does not fully dissociate) and m = 0.05. So, delta T = 1 * K * 0.05 = 0.05K.
Now, we can arrange the solutions in increasing order of delta T:
0.05 m CH3COOH < 0.15 m C6H12O6 < 0.35 m NaCl < 0.20 m Na3PO4 < 0.30 m MgCl2